3.5.18 \(\int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [418]

Optimal. Leaf size=36 \[ \frac {2 i \sqrt {e \sec (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

2*I*(e*sec(d*x+c))^(1/2)/d/(a+I*a*tan(d*x+c))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {3569} \begin {gather*} \frac {2 i \sqrt {e \sec (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Sec[c + d*x]]/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((2*I)*Sqrt[e*Sec[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3569

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {2 i \sqrt {e \sec (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 36, normalized size = 1.00 \begin {gather*} \frac {2 i \sqrt {e \sec (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[e*Sec[c + d*x]]/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((2*I)*Sqrt[e*Sec[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (30 ) = 60\).
time = 0.88, size = 74, normalized size = 2.06

method	result	size
risch	\(\frac {2 i \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}{\sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(59\)
default	\(\frac {2 i \sqrt {\frac {e}{\cos \left (d x +c \right )}}\, \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )}{d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right ) a}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*I/d*(e/cos(d*x+c))^(1/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)/(I*sin(d*x+c)+cos(d*x+c))
/a

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (28) = 56\).
time = 0.49, size = 75, normalized size = 2.08 \begin {gather*} \frac {2 i \, \sqrt {-\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1} e^{\frac {1}{2}}}{\sqrt {a} d \sqrt {-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2*I*sqrt(-sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)*e^(1/2)/(sqrt(a)*d*sqrt(-2*I*sin(d*x + c)/(cos(d*x + c) + 1
) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1))

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (28) = 56\).
time = 0.35, size = 64, normalized size = 1.78 \begin {gather*} \frac {2 \, \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\frac {1}{2}} + i \, e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{a d \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(I*e^(1/2) + I*e^(2*I*d*x + 2*I*c + 1/2))*e^(-1/2*I*d*x - 1/2*I*c)/(a*d*sq
rt(e^(2*I*d*x + 2*I*c) + 1))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {e \sec {\left (c + d x \right )}}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(1/2)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(e*sec(c + d*x))/sqrt(I*a*(tan(c + d*x) - I)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(e^(1/2)*sqrt(sec(d*x + c))/sqrt(I*a*tan(d*x + c) + a), x)

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Mupad [B]
time = 4.31, size = 40, normalized size = 1.11 \begin {gather*} \frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,2{}\mathrm {i}}{d\,\sqrt {a+\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{\cos \left (c+d\,x\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(1/2)/(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

((e/cos(c + d*x))^(1/2)*2i)/(d*(a + (a*sin(c + d*x)*1i)/cos(c + d*x))^(1/2))

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